LeetCode #2132 — HARD

Stamping the Grid

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an m x n binary matrix grid where each cell is either 0 (empty) or 1 (occupied).

You are then given stamps of size stampHeight x stampWidth. We want to fit the stamps such that they follow the given restrictions and requirements:

Cover all the empty cells.
Do not cover any of the occupied cells.
We can put as many stamps as we want.
Stamps can overlap with each other.
Stamps are not allowed to be rotated.
Stamps must stay completely inside the grid.

Return true if it is possible to fit the stamps while following the given restrictions and requirements. Otherwise, return false.

Example 1:

Input: grid = [[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0]], stampHeight = 4, stampWidth = 3
Output: true
Explanation: We have two overlapping stamps (labeled 1 and 2 in the image) that are able to cover all the empty cells.

Example 2:

Input: grid = [[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]], stampHeight = 2, stampWidth = 2 
Output: false 
Explanation: There is no way to fit the stamps onto all the empty cells without the stamps going outside the grid.

Constraints:

m == grid.length
n == grid[r].length
1 <= m, n <= 10⁵
1 <= m * n <= 2 * 10⁵
grid[r][c] is either 0 or 1.
1 <= stampHeight, stampWidth <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an m x n binary matrix grid where each cell is either 0 (empty) or 1 (occupied). You are then given stamps of size stampHeight x stampWidth. We want to fit the stamps such that they follow the given restrictions and requirements: Cover all the empty cells. Do not cover any of the occupied cells. We can put as many stamps as we want. Stamps can overlap with each other. Stamps are not allowed to be rotated. Stamps must stay completely inside the grid. Return true if it is possible to fit the stamps while following the given restrictions and requirements. Otherwise, return false.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0],[1,0,0,0]]
4
3

Example 2

[[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]]
2
2

Core Insight

What unlocks the optimal approach

Can we use prefix sums here?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2132: Stamping the Grid
class Solution {
    public boolean possibleToStamp(int[][] grid, int stampHeight, int stampWidth) {
        int m = grid.length, n = grid[0].length;
        int[][] s = new int[m + 1][n + 1];
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + grid[i - 1][j - 1];
            }
        }
        int[][] d = new int[m + 2][n + 2];
        for (int i = 1; i + stampHeight - 1 <= m; ++i) {
            for (int j = 1; j + stampWidth - 1 <= n; ++j) {
                int x = i + stampHeight - 1, y = j + stampWidth - 1;
                if (s[x][y] - s[x][j - 1] - s[i - 1][y] + s[i - 1][j - 1] == 0) {
                    d[i][j]++;
                    d[i][y + 1]--;
                    d[x + 1][j]--;
                    d[x + 1][y + 1]++;
                }
            }
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                d[i][j] += d[i - 1][j] + d[i][j - 1] - d[i - 1][j - 1];
                if (grid[i - 1][j - 1] == 0 && d[i][j] == 0) {
                    return false;
                }
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #2132: Stamping the Grid
func possibleToStamp(grid [][]int, stampHeight int, stampWidth int) bool {
	m, n := len(grid), len(grid[0])
	s := make([][]int, m+1)
	for i := range s {
		s[i] = make([]int, n+1)
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + grid[i-1][j-1]
		}
	}

	d := make([][]int, m+2)
	for i := range d {
		d[i] = make([]int, n+2)
	}

	for i := 1; i+stampHeight-1 <= m; i++ {
		for j := 1; j+stampWidth-1 <= n; j++ {
			x, y := i+stampHeight-1, j+stampWidth-1
			if s[x][y]-s[x][j-1]-s[i-1][y]+s[i-1][j-1] == 0 {
				d[i][j]++
				d[i][y+1]--
				d[x+1][j]--
				d[x+1][y+1]++
			}
		}
	}

	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			d[i][j] += d[i-1][j] + d[i][j-1] - d[i-1][j-1]
			if grid[i-1][j-1] == 0 && d[i][j] == 0 {
				return false
			}
		}
	}
	return true
}

# Accepted solution for LeetCode #2132: Stamping the Grid
class Solution:
    def possibleToStamp(
        self, grid: List[List[int]], stampHeight: int, stampWidth: int
    ) -> bool:
        m, n = len(grid), len(grid[0])
        s = [[0] * (n + 1) for _ in range(m + 1)]
        for i, row in enumerate(grid, 1):
            for j, v in enumerate(row, 1):
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + v
        d = [[0] * (n + 2) for _ in range(m + 2)]
        for i in range(1, m - stampHeight + 2):
            for j in range(1, n - stampWidth + 2):
                x, y = i + stampHeight - 1, j + stampWidth - 1
                if s[x][y] - s[x][j - 1] - s[i - 1][y] + s[i - 1][j - 1] == 0:
                    d[i][j] += 1
                    d[i][y + 1] -= 1
                    d[x + 1][j] -= 1
                    d[x + 1][y + 1] += 1
        for i, row in enumerate(grid, 1):
            for j, v in enumerate(row, 1):
                d[i][j] += d[i - 1][j] + d[i][j - 1] - d[i - 1][j - 1]
                if v == 0 and d[i][j] == 0:
                    return False
        return True

// Accepted solution for LeetCode #2132: Stamping the Grid
impl Solution {
    pub fn possible_to_stamp(grid: Vec<Vec<i32>>, stamp_height: i32, stamp_width: i32) -> bool {
        let n: usize = grid.len();
        let m: usize = grid[0].len();

        let mut prefix_vec: Vec<Vec<i32>> = vec![vec![0; m + 1]; n + 1];

        // Initialize the prefix vector
        for i in 0..n {
            for j in 0..m {
                prefix_vec[i + 1][j + 1] =
                    prefix_vec[i][j + 1] + prefix_vec[i + 1][j] - prefix_vec[i][j] + grid[i][j];
            }
        }

        let mut diff_vec: Vec<Vec<i32>> = vec![vec![0; m + 1]; n + 1];

        // Construct the difference vector based on prefix sum vector
        for i in 0..n {
            for j in 0..m {
                // If the value of the cell is one, just bypass this
                if grid[i][j] == 1 {
                    continue;
                }
                // Otherwise, try stick the stamp
                let x: usize = i + (stamp_height as usize);
                let y: usize = j + (stamp_width as usize);
                // Check the bound
                if x <= n && y <= m {
                    // If the region can be sticked (All cells are empty, which means the sum will be zero)
                    if prefix_vec[x][y] - prefix_vec[x][j] - prefix_vec[i][y] + prefix_vec[i][j]
                        == 0
                    {
                        // Update the difference vector
                        diff_vec[i][j] += 1;
                        diff_vec[x][y] += 1;

                        diff_vec[x][j] -= 1;
                        diff_vec[i][y] -= 1;
                    }
                }
            }
        }

        let mut check_vec: Vec<Vec<i32>> = vec![vec![0; m + 1]; n + 1];

        // Check the prefix sum of difference vector, to determine if there is any empty cell left
        for i in 0..n {
            for j in 0..m {
                // If the value of the cell is one, just bypass this
                if grid[i][j] == 1 {
                    continue;
                }
                // Otherwise, check if the region is empty, by calculating the prefix sum of difference vector
                check_vec[i + 1][j + 1] =
                    check_vec[i][j + 1] + check_vec[i + 1][j] - check_vec[i][j] + diff_vec[i][j];
                if check_vec[i + 1][j + 1] == 0 {
                    return false;
                }
            }
        }

        true
    }
}

// Accepted solution for LeetCode #2132: Stamping the Grid
function possibleToStamp(grid: number[][], stampHeight: number, stampWidth: number): boolean {
    const m = grid.length;
    const n = grid[0].length;
    const s: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + grid[i - 1][j - 1];
        }
    }

    const d: number[][] = Array.from({ length: m + 2 }, () => Array(n + 2).fill(0));
    for (let i = 1; i + stampHeight - 1 <= m; ++i) {
        for (let j = 1; j + stampWidth - 1 <= n; ++j) {
            const [x, y] = [i + stampHeight - 1, j + stampWidth - 1];
            if (s[x][y] - s[x][j - 1] - s[i - 1][y] + s[i - 1][j - 1] === 0) {
                d[i][j]++;
                d[i][y + 1]--;
                d[x + 1][j]--;
                d[x + 1][y + 1]++;
            }
        }
    }

    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            d[i][j] += d[i - 1][j] + d[i][j - 1] - d[i - 1][j - 1];
            if (grid[i - 1][j - 1] === 0 && d[i][j] === 0) {
                return false;
            }
        }
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m × n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.